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字符串 S 由小写字母组成。我们要把这个字符串划分为尽可能多的片段，同一个字母只会出现在其中的一个片段。返回一个表示每个字符串片段的长度的列表。


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        <h1 class="title">leetcode763.划分字母区间</h1>
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            <span>二月 23, 2020</span>
            

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            <h1 id="leetcode763-划分字母区间"><a href="#leetcode763-划分字母区间" class="headerlink" title="leetcode763. 划分字母区间"></a>leetcode763. 划分字母区间</h1><hr>
<blockquote>
<p>字符串 S 由小写字母组成。我们要把这个字符串划分为尽可能多的片段，同一个字母只会出现在其中的一个片段。返回一个表示每个字符串片段的长度的列表。</p>
</blockquote>
<blockquote>
<p>示例 1:<br>输入: S = “ababcbacadefegdehijhklij”<br>输出: [9,7,8]<br>解释:<br>划分结果为 “ababcbaca”, “defegde”, “hijhklij”。<br>每个字母最多出现在一个片段中。<br>像 “ababcbacadefegde”, “hijhklij” 的划分是错误的，因为划分的片段数较少。</p>
</blockquote>
<h3 id="我的思路："><a href="#我的思路：" class="headerlink" title="我的思路："></a>我的思路：</h3><p>用一个map来记录字母出现的最后一次的位置，记做last。然后遍历这之前的所有字母，检查是否有值比这个last大，比这个大，说明这个字母不能作为分隔点。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> List&lt;Integer&gt; <span class="title">partitionLabels</span><span class="params">(String S)</span> </span>&#123;</span><br><span class="line">      LinkedHashMap&lt;Character,Integer&gt;map=<span class="keyword">new</span> LinkedHashMap&lt;&gt;();</span><br><span class="line">       <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;S.length();i++)&#123;</span><br><span class="line">           	map.put(S.charAt(i), i);</span><br><span class="line">       &#125;</span><br><span class="line">       <span class="comment">//如果用linkedHashMap下面这些排序的过程就能省略</span></span><br><span class="line">		<span class="comment">// List&lt;Map.Entry&lt;Character,Integer&gt;&gt;listEntry=new ArrayList&lt;&gt;();</span></span><br><span class="line">		<span class="comment">// listEntry.addAll(map.entrySet());</span></span><br><span class="line">		<span class="comment">// Collections.sort(listEntry, new Comparator&lt;Map.Entry&lt;Character, Integer&gt;&gt;()&#123;</span></span><br><span class="line">		<span class="comment">// @Override</span></span><br><span class="line">		<span class="comment">// public int compare(Map.Entry&lt;Character, Integer&gt; o1, Map.Entry&lt;Character, Integer&gt; o2) &#123;</span></span><br><span class="line">		<span class="comment">// 	// TODO Auto-generated method stub</span></span><br><span class="line">		<span class="comment">// 	return o1.getValue()-o2.getValue();</span></span><br><span class="line">		<span class="comment">// &#125;</span></span><br><span class="line">		<span class="comment">// &#125;);</span></span><br><span class="line">       ArrayList&lt;Integer&gt;result=<span class="keyword">new</span> ArrayList&lt;&gt;();</span><br><span class="line">       <span class="keyword">int</span> last=<span class="number">0</span>,flag=<span class="number">0</span>;</span><br><span class="line">       <span class="keyword">for</span>(Map.Entry&lt;Character,Integer&gt;entry:map.entrySet()) &#123;</span><br><span class="line">       	flag=<span class="number">0</span>;</span><br><span class="line">       	<span class="keyword">char</span> nowChar=entry.getKey();</span><br><span class="line">       	<span class="keyword">int</span> now=entry.getValue();</span><br><span class="line">           <span class="keyword">if</span>(last&gt;now)<span class="keyword">continue</span>;<span class="comment">//linkedHashMap才需要比较</span></span><br><span class="line">       	<span class="keyword">for</span>(<span class="keyword">int</span> i=last;i&lt;now;i++) &#123;</span><br><span class="line">       		<span class="keyword">if</span>(map.get(S.charAt(i))&gt;now) &#123;</span><br><span class="line">       			flag=<span class="number">1</span>;</span><br><span class="line">       			<span class="keyword">break</span>;</span><br><span class="line">       		&#125;</span><br><span class="line">       	&#125;</span><br><span class="line">       	<span class="keyword">if</span>(flag==<span class="number">1</span>)<span class="keyword">continue</span>;</span><br><span class="line">       	result.add(now-last+<span class="number">1</span>);</span><br><span class="line">       	last=now+<span class="number">1</span>;</span><br><span class="line">       &#125;</span><br><span class="line">       <span class="keyword">return</span> result;</span><br><span class="line">   &#125;</span><br></pre></td></tr></table></figure>
<p>我自己的写法中在for内部又多了一层for，这样浪费了时间，看看官方解法：</p>
<h3 id="官方解法"><a href="#官方解法" class="headerlink" title="官方解法"></a>官方解法</h3><p>维护一个窗口，如果当前字母出现的最后一次的位置（last）比右边界大，那么需要拓展窗口，当 <strong>i</strong> 与 <strong>j</strong> 相遇的时候,说明窗口内部所有的字母都符合条件。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> List&lt;Integer&gt; <span class="title">partitionLabels</span><span class="params">(String S)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span>[] last = <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">26</span>];</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; S.length(); ++i)</span><br><span class="line">            last[S.charAt(i) - <span class="string">'a'</span>] = i;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">int</span> j = <span class="number">0</span><span class="comment">//右窗口;</span></span><br><span class="line">        <span class="keyword">int</span> anchor = <span class="number">0</span><span class="comment">//左窗口;</span></span><br><span class="line">        List&lt;Integer&gt; ans = <span class="keyword">new</span> ArrayList();</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; S.length(); ++i) &#123;</span><br><span class="line">            j = Math.max(j, last[S.charAt(i) - <span class="string">'a'</span>]);</span><br><span class="line">            <span class="keyword">if</span> (i == j) &#123;</span><br><span class="line">                ans.add(j - anchor + <span class="number">1</span>);</span><br><span class="line">                anchor = i + <span class="number">1</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>
<p><strong>leetcode 21/100</strong></p>

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